Bonjour rationnelle!

Let circle $C$ have the equation $x^2 + (y-\sqrt{2})^2 = r^2$ , or $x^2 + (y^2 - 2\sqrt{2}y + 2) = r^2.$ Clearly, $y=0$ is a necessary & sufficient condition to ensure $C$ contains rational points. This reduces our circular equation down to:

$x^2 + 2 = r^2 \Rightarrow 2 = r^2-x^2 = (r+x)(r-x)$

which further necessitates $r \ge \sqrt{2}$ . If $r = \sqrt{2}$ , then $C$ only has the rational point $(x,y) = (0,0)$ . If $r > \sqrt{2}$ , suppose we let:

$\large r+x=n, r-x=\frac{2}{n}$

for $n \in \mathbb{Q}$ . Solving this system of equations ultimately yields:

$\large r = \frac{n^2+2}{2n}, x = \frac{n^2-2}{2n}$

This also includes $\large x = -\frac{n^2-2}{2n}$ since $C$ is symmetric about the $y-$ axis $\Rightarrow 2$ rational points. Hence, the maximum number of rational points $k \in C$ is $\boxed{ k \le 2}.$